Small cancellation theory and Burnside problem.

Transcription

1 Small cancellation theory and Burnside problem. Rémi Coulon February 27, 2013 Abstract In these notes we detail the geometrical approach of small cancellation theory used by T. Delzant and M. Gromov to provide a new proof of the infiniteness of free Burnside groups and periodic quotients of torsion-free hyperbolic groups. Contents 1 Introduction Usual small cancellation theory Small cancellation theory in hyperbolic groups Iterating small cancellation Outline of the paper Hyperbolic geometry The four points inequality Quasi-geodesics Quasi-convex subsets Ultra-limit of hyperbolic spaces Isometries of a hyperbolic space Group acting on a hyperbolic space Rotation families Fundamental theorem Consequences Cone over a metric space Definition and metric Hyperbolicity of a cone Group action on a cone Cone-off construction Metric on the cone-off Uniform approximation of the distance Ultra-limit and cone-off Hyperbolicity of the cone-off

2 2 Small cancellation theory and Burnside problem. 6 Small cancellation theory General framework Isometries of the quotient Groups without even-torsion Applications Periodic quotients of hyperbolic groups A few words about Gromov s monster group A Appendix: Cartan-Hadamard Theorem 58 A.1 Following paths. Definition and first properties A.2 Transitivity of the relation A.3 Existence of following quasi-geodesics A.4 The space of quasi-geodesic paths A.5 Global hyperbolicity of X Introduction Let n be an integer. A group G has exponent n if for all g G, g n = 1. In 1902 W. Burnside asked wether a finitely generated group with finite exponent is necessarily finite or not [5]. To study this question it is natural to look at the free Burnside group B r (n = F r /F r n, which is the quotient of the free group of rank r by the (normal subgroup generated by the n-th powers of all elements. It is indeed the largest group of rank r and exponent n. In 1968, P. S. Novikov and S. I. Adian answered negatively the Burnside Problem. They proved that if r 2 and n is odd larger than 4381, then B r (n is infinite [19]. Since, this result has been extended in many directions (even exponents, periodic quotient of hyperbolic groups, etc [21, 22, 15, 18]. More recently T. Delzant and M. Gromov provided a new approach of the Burnside problem [10]. In particular, they give an alternative proof of the following theorem. Theorem 1.1 ([10, Th ], see also [22]. Let G be a non-cyclic, torsion-free hyperbolic group. There exists an integer n 0 such that for all odd exponents n n 0, the quotient G/G n is infinite. The aim of these notes is to give a comprehensive presentation of their method. Actually a more general statement holds for an arbitrary hyperbolic group. Theorem 1.2 ([16, Th. A]. Let G be a non-virtually cyclic hyperbolic group. For every integer n 0 there exists an exponent n n 0 such that the quotient G/G n is infinite. However this second result require to control the even torsion, which is much harder. The notes do not cover this theorem. From a geometrical point of view the difficulty to study Burnside groups comes from the fact that we have no nice metric space on which B r (n is acting by isometries. Therefore the idea is to study B r (n as the direct limit of a sequence of groups, F r = G 0 G 1 G 3 G 4 G k G k+1... where each G k is easier to understand. In this construction, T. Delzant and M. Gromov obtain G k+1 as a quotient of G k using a geometrical version of the small cancellation theory. It forces the groups G k to be non-virtually cyclic and hyperbolic. Thus their limit B r (n cannot be finite. All

3 1 Introduction 3 the other known strategies also use such a sequence of groups. Before explaining the construction of T. Delzant and M. Gromov let us recall a few ideas about the usual small cancellation theory and its geometric generalization given by M. Gromov [13]. 1.1 Usual small cancellation theory. For more details about the usual small cancellation theory, we refer the reader to [17]. Let F(S be the free group generated by a finite set S. Let R be a set of words over the alphabet S S 1. The goal is to study the group Ḡ = F(S/ R, where R stands for the normal subgroup of F(S generated by R. We assume that the elements of R are non-trivial and cyclically reduced. We denote by R the set of all cyclic conjugates of elements of R R 1. A piece is a common prefix of two distinct elements of R. In other words a piece is a subword that could cancel in the product rs where r, s R. Let λ > 0. One says that R satisfies the small cancellation assumption C (λ if for all pieces u, for all relations r R containing u, u λ r. Let us mention one important theorem. In the next paragraph we will provide an extension of it. Theorem 1.3 (see [11]. Let λ (0, 1/6. Let R be a set of non-trivial, cyclically reduced words over the alphabet S S 1. If R satisfies the condition C (λ then the quotient F(S/ R is a hyperbolic group. Four our purpose we are going to consider a more stronger condition called C (λ. It requires that for all pieces u, for all relations r R (not necessarily containing u, u λ r. This assumption can be reformulated in a more geometrical way. To that end we consider the Cayley graph of F(S with respect to S, denoted by X which is a tree. Let r R. It fixes two points r and r + of the boundary at infinity X of X. These points are joined by a bi-infinite geodesic called the axis of r. We denote it by Y r. The isometry r acts on Y r by translation of length [r] = inf x X rx x, where x y denotes the distance between x and y. Since r is a cyclically reduced word, [r] is in fact the length of the relation r. Consider now two relations r, s R. The length of a common piece of r and s is exactly diam (Y r Y s. Thus the C (λ condition can be stated in this way: diam (Y r Y s λ inf [r]. r sup r s With this idea in mind we provide in the next paragraph a larger framework to the small cancellation theory. 1.2 Small cancellation theory in hyperbolic groups From now on X is a proper geodesic δ-hyperbolic space. Let G be a group acting properly co-compactly by isometries on X. The translation length of an isometry g G, denoted by [g] is the quantity [g] = inf x X gx x. We consider a collection Q of pairs (H, Y satisfying the following properties (i Y is a 2δ-quasi-convex subset of X and H a subgroup of Stab (Y, the stabilizer in G of Y, acting co-compactly on Y. (ii the family Q is invariant under the action of G i.e., for all g G, for all (H, Y Q, (ghg 1, gy is an element of Q, (iii the set Q/G is finite. Our goal is to study the group Ḡ = G/K, where K is the (normal subgroup of G generated by the H s. For this purpose, we define two quantities which respectively play the role of the length of the largest piece and the length of the smallest relation.

4 4 Small cancellation theory and Burnside problem. The maximal overlap between two quasi-convex of Q is measured by the quantity (Q = sup diam ( Y 1 +5δ Y +5δ 2. (H 1,Y 1 (H 2,Y 2 The smallest translation length of Q is defined as T (Q = inf {[h] h H \ {1}, (H, Y Q}. Note that if (Q is finite then two distinct groups H cannot be associated in Q to the same quasi-convex Y. In particular H is a normal subgroup of Stab (Y. We are interested in situations where the ratios (Q/T (Q and δ/t (Q are very small (see Theorem 1.4. In particular it forces every non-trivial element of every subgroup H to be hyperbolic. Examples. (i Let R be a finite set of elements of F(S. Consider the collection Q = { ( uru 1, uy r r R, u F(S }. In this case we recover the usual small cancellation theory. Since the the Cayley graph of F(S is 0-hyperbolic, the assumption C (λ is equivalent to (Q λt (Q (ii The next example comes from small cancellation with graphs (see [20]. Let Γ be a finite connected graph whose edges are labelled with elements of S S 1. Given a vertex v 0 of Γ, the labeling provides a homomorphism from the fundamental group of (Γ, v 0 onto a subgroup H of F(S. Moreover, we have a map from Γ, the universal cover of Γ, onto a subtree T of the Cayley graph of F(S. We can consider the collection Q = { ( ghg 1, gt g G/H }. In this situation a piece is, by definition, a word which labels two distinct simple paths of Γ. Then (Q is the length of the largest piece whereas T (Q is the girth of the graph Γ i.e., the length of the smallest embedded loop in Γ. For further detail we refer the reader to [14, 1/6-theorem] and [20, Th.1]. (iii Let G be a group acting properly, co-compactly by isometries on a geodesic, δ-hyperbolic space X. If r G is a hyperbolic isometry it fixes two points r and r + of X, the boundary at infinity of X. Following the case of free groups we define Y r to be the set of all points of X which are 10δ-close to some bi-infinite geodesic joining r and r +. This set is 2δ-quasi-convex. We consider the set P of all hyperbolic elements r G which are not proper powers such that [r] 1000δ. To every integer n we associate the collection Q n = { ( r n, Y r r P } Assume now that G is non-virtually cyclic. Since G acts properly co-compactly on X, the overlap (Q n is uniformly bounded. Therefore, by choosing an integer n large enough, one can obtain a ratio (Q n /T (Q n as small as desired.

5 1 Introduction 5 Consider now an arbitrary family Q as defined previously. In order to study the quotient Ḡ = G/K we construct a space X on which Ḡ acts properly, co-compactly, by isometries. Roughly speaking, X is obtained as follows. We fix a large real number ρ. Its value will be made precise later. For each (H, Y Q, we start by attaching on X along Y a cone of base Y and radius ρ. The cones are endowed with a negatively curved metric modeled on the distance of the hyperbolic plane H 2. This new space Ẋ (called cone-off of X inherits a metric coming from the distance on X and the ones on the cones. Since Q is endowed with an action of G, the action of G on X extends by homogeneity into an action of G on Ẋ. On this space every group H acts as a rotation of very large angle that fixes the apex of the cone of base Y. One speaks of rotation family (see Section 3. The space X is then the quotient of Ẋ by K. By construction X is proper and geodesic. Moreover, Ḡ acts properly co-compactly by isometries on it. The study of the metric on Ẋ and X is a key question. This will be done in Sections 4-6. One of the main features of the space X is the following. Assume that the ratio (Q/T (Q is very small, then every ball of X of radius ρ/50 is roughly δ-hyperbolic where δ is the hyperbolicity constant of the hyperbolic plane H 2. If ρ has been chosen large enough, we can then apply a Cartan-Hadamard Theorem (see Theorem A.1. It leads to the following analogue of Theorem 1.3. Theorem 1.4. There exist positive constants δ 0 and λ 0 (which do not depend on X, G or Q with the following property. Assume that δ T (Q δ 0 and (Q T (Q λ 0. Then X is a hyperbolic space endowed with a proper co-compact action of Ḡ. Some hypotheses of the theorem can be weaken. For instance we do not really need that the action of G (respectively H on X (respectively Y to be co-compact. Without these assumptions, the space X remains hyperbolic nevertheless the action of Ḡ on X is no more proper or co-compact. 1.3 Iterating small cancellation Theorem 1.4 is an important step of the induction process to prove the infiniteness of the Burnside groups. However this is definitely not sufficient. The space X has in fact many other properties that we will enlighten in Section 6. Let us explain briefly where the difficulty does come from. Let G 0 be a non-virtually cyclic torsion-free group acting properly co-compactly on a δ-hyperbolic space X 0 and n 0 an integer. Following Example (iii, we consider the quotient G 1 = G 0 /K 0 where K 0 is the normal subgroup of G 0 generated by {r n0 /r G, hyperbolic, not a proper power, [r] 1000δ} If n 0 is large enough, Theorem 1.4 applies. The quotient G 1 is also a hyperbolic group. Thus we would like to iterate the process and kill some large powers in G 1. According to Example (iii this can be done, but there is no reason that the exponent n 1 that works for the second step should be the same as n 0. In this way one can construct by induction an infinite torsion group. Nevertheless to study Burnside groups we need to follow during the construction some parameters that allow us to use at each step the same exponent n. Note already that the constants δ 0 and λ 0 in Theorem 1.4 do not depend on G, X or Q. This is an important but not sufficient fact. To control the ratio (Q/T (Q we will take care of two quantities associated to the action of G on X. The radius of injectivity which represent the smallest asymptotic translation length of a

6 6 Small cancellation theory and Burnside problem. hyperbolic element of G and the invariant A(G, X which measures the maximal overlap between the axes of two small isometries of G (see Section 2.6 for the precise definitions. During this construction we will require the exponent n to be odd. With this assumption we can prove that every subgroup of any G k is either cyclic or contains a free group. It will help to control (Q in terms of A(G, X. Burnside groups of large even exponents are known to be infinite [15, 18]. However the proof is much harder. The hypotheses required to apply small cancellation are not very restricting. See for instance [23]. Therefore iterating small cancellation can be exploited to build many groups with pathological properties. It is one of the ingredients involved in the construction of Gromov s monster group [14, 1]. This object is a limit of groups G k where G k+1 is obtained from G k using graphical small cancellation (see Example (ii. One remarkable property of this group is that it does not coarsely embed into a Hilbert space and thus does not satisfy the Baum-Connes conjecture with coefficients. 1.4 Outline of the paper In the Section 2 we recall the main features of hyperbolic spaces in the sense of Gromov and groups acting on a hyperbolic space. Section 3 is dedicated to the study of rotation families. This section differs from the tools used by T. Delzant and M. Gromov. In this way we do not need to use orbifolds as in [10]. Sections 4 and 5 detail the cone-off construction. The core of the proof is contained in Section 6. We prove among others the small cancellation theorem (see Theorem 6.11 and investigate the properties of the space X. Thanks to the small cancellation we state at the beginning of Section 7.1 an induction lemma which explains how the group G k+1 can be obtained from G k. In particular it provides a set of assumptions which if they are satisfied for G k also hold for G k+1. This allow us to iterate the construction and prove the main theorem (see Theorem 7.7. Our approach of the small cancellation theory is very broad. In particular it covers all the main results that are needed to construct Gromov s monster group. In Section 7.2 we also give some keys to understand this construction as explained in [1]. In the appendix, we propose an alternative proof of the Cartan-Hadamard Theorem based on the ideas given in [10]. Acknowledgments. These notes follow a series of lectures given at the Vanderbilt University during Fall The author wishes to express his gratitude to M. Mihalik, A. Ol shanskiĭ, D. Osin and M. Sapir for their patient and active interest in this work and also for encouraging him to write this text. He is also greatly thankful to T. Delzant who guides his first steps in this field. 2 Hyperbolic geometry In this section we some of the standard facts on hyperbolic spaces in the sense of M. Gromov. We only give the proofs of quantitative results, which are not a straight forward application of the four points condition. For more details we refer the reader to the original paper of M. Gromov [12] or to [6, 11]. Let X be a metric length space. Unless otherwise stated a path is a rectifiable path parametrized by arclength. Given two points x and x of X, we denote by x x X (or simply x x the distance between them. We write B(x, r for the closed ball of center x and radius r i.e., the set of points y X such that x y r. Let Y be a subset of X. The distance between a point

7 2 Hyperbolic geometry 7 x of X and Y is denoted by d(x, Y. We write Y +α for the α-neighborhood of Y i.e., the set of points x X such that d(x, Y α. Let η 0. A point p of Y is an η-projection of x X on Y if x p d(x, Y + η. A 0-projection is simply called a projection. 2.1 The four points inequality. The Gromov product of three points x, y, z X is defined by x, y z = 1 2 { } x z + y z x y. The space X is δ-hyperbolic if for every x, y, z, t X } x, z t min { x, y t, y, z t δ, (1 or equivalently { } x z + y t max x y + z t, y z + x t + 2δ. (2 To show that a space is hyperbolic, it is actually sufficient to prove (1 with a fixed base point. Lemma 2.1 (see [6, Chap.1 Prop. 1.2]. Let δ 0. Let (X, t be a pointed metric space. If for every x, y, z X, we have x, z t min { x, y t, y, z t } δ, then X is 2δ-hyperbolic. Remarks. Note that in the definition of hyperbolicity we do not assume that X is a geodesic. In Section 5 we construct a length space (the cone-off for which it is not obvious to prove that it is geodesic. However it satisfies (1. Therefore we prefer to state all the results concerning hyperbolicity for the class of length spaces. To compensate for the absence of geodesics we use the following property. For every x, x X, for every η > 0, for every 0 l x x there exists y X such that x y = l and x, x y η. In this context, the Gromov product x, x y should be thought as an analogue of the distance between y and a geodesic joining x and x. If X is 0-hyperbolic, then it can be isometrically embedded in an R-tree [11, Chap. 2, Prop. 6]. However we will always assume that the hyperbolicity constant δ is positive. Most of the results hold for δ = 0. But this is more convenient to define particular subsets (see for instance Definition 2.14 or Lemmas 2.12 and 2.13 without introducing other auxiliary positive parameters. The hyperbolicity constant of the hyperbolic plane H 2 will play a particular role. Therefore we denote it by δ (bold delta. From now on we assume that the space X is δ-hyperbolic. It is known that triangles in a hyperbolic geodesic space are thin (every side lies in a uniform neighborhood of the union of the two others. Since our space is not necessarily geodesic, we use instead the following metric inequalities. They will be used in situations where the Gromov products x, y s, x, y t or x, z t are small. Their proof is left to the reader. Lemma 2.2. Let x, y, z, s and t be five points of X. } (i x, y t max { x t y, z x, x, z t + δ,

8 8 Small cancellation theory and Burnside problem. } (ii s t x s x t + 2 max { x, y s, x, y t + 2δ, (iii The distance s t is bounded above by { } x max s x t + 2 max { x, y s, x, z t }, x s + x t 2 y, z x + 4δ. 2.2 Quasi-geodesics Definition 2.3. Let l 0, k 1 and L 0. Let I be an interval of R. A path γ : I X is (i a (k, l-quasi-geodesic if for all t, t I γ(t γ(t t t k γ(t γ(t + l. (ii an L-local (k, l-quasi-geodesic if the restriction of γ to any interval of length L is a (k, l- quasi-geodesic. Remarks. The first inequality in the definition just follows from the fact that γ is parametrized by arc length. Since X is a length space, for every x, x X, for every l > 0, there exists a (1, l- quasi-geodesic joining x and x (take a rectifiable path between them whose length is shorter than x x + l. Let γ : I X be a (1, l-quasi-geodesic. Let x = γ(t, x = γ(t and y = γ(s be three points on γ. If t s t then x, x y l/2. Proposition 2.4. Let γ : I X be a (1, l-quasi-geodesic of X. (i Let x be a point of X and p an η-projection of x on γ. For all y γ, x, y p l + η + 2δ. (ii For every x X for every y, y γ we have y, y x l d(x, γ y, y x + l + 3δ. Proof. Let x be a point of X. We denote by p = γ(s an η-projection of x on γ and y = γ(t a point of γ. By reversing if necessary the parametrization of γ we can assume that s t. Note that y p x, y p. Therefore there exists r [s, t] such that the point z = γ(r satisfies z p = x, y p and p, y z l/2. By Lemma 2.2-(i we have x, p z l/2 + δ i.e., x, y p = p z x p x z + l + 2δ. Nevertheless, p is an η-projection of x on γ. Thus x p x z η. It follows that x, y p l + η + 2δ, which proves (i. Let x X. Let y and y be two points of γ. The first inequality of (ii follows from the triangle inequality. We denote by p a projection of x on γ. It follows from Point (i that d(x, γ = y, p x + x, y p is bounded above by y, p x + l + 2δ. In the same way d(x, γ y, p x + l + 2δ. Consequently, the hyperbolicity condition (1 gives d(x, γ min { y, p x, y, p x } + l + 2δ y, y x + l + 3δ. Proposition 2.5 (Stability of quasi-geodesics [6, Chap. 3, Th ]. Let k 1, k > k and l 0. There exists L 0, D 0 which only depend on δ, k, k and l such that the Hausdorff distance between two L-local (k, l-quasi-geodesics joining the same endpoints is at most D. Moreover, every L-local (k, l-quasi-geodesic is also a (global (k, l-quasi-geodesic.

9 2 Hyperbolic geometry 9 In this paper we are mostly using L-local (1, l-quasi-geodesics. For these paths one can easily provide a precise value for D (see Corollary 2.6. This is not really crucial but it will decrease the number of parameters that we have to deal with in all the proofs. Corollary 2.6. Let l, l 0. There exists L = L(l, l, δ which only depends on δ, l and l with the following property. Let γ : [a, b] X be a L-local (1, l-quasi-geodesic and γ : [a, b ] X a L-local (1, l -quasi-geodesic. If they join the same extremities then γ lies in the (l/2 + l + 5δ- neighborhood of γ. Proof. Let l 0. The constants L, D and k are chosen so that we can apply Proposition 2.5 to γ and γ : the Hausdorff distance between γ and γ is at most D and γ is a (k, l -global quasi-geodesic. Note that k, D and L only depend on l, l and δ. Without loss of generality we may assume that L > k(4d + 3l + 6δ + l. Let x = γ(t be a point on γ. We assume that a t and b t are bounded below by D + 3l/2 + 3δ. The other cases are similar. Let us put t ± = t ± (D + 3l/2 + 3δ and x ± = γ(t ±. By the stability of quasi-geodesics there exist points y ± = γ (s ± such that x ± y ± D. Without loss of generality we can assume that s s +. By hyperbolicity we obtain min { x x x y, y, y + x, x + x x + y + } x, x + x + 2δ l/2 + 2δ. However by construction x ± x is bounded below by D + l/2 + 3δ and x ± y ± above by D. Therefore, we necessarily have y, y + x l/2 + 2δ. We now claim that s + s L. Indeed γ is a (k, l -quasi-geodesic. Therefore the triangle inequality leads to k 1 ( s + s l y + y x + x + 2D 4D + 3l + 6δ Our claim follows from the assumption on L. In particular γ restricted to [s, s + ] is a (global (1, l-quasi-geodesic. By Proposition 2.4-(ii, the point x is (l/2 + l + 5δ-close to γ. Remarks. We keep the notations of the corollary. Let x = γ(t, x = γ(t and y = γ(s be three points on γ. If t s t then x, x y l/2 + 5δ. Moreover, for every p X we have d(p, γ x, x p + l + 8δ. Proposition 2.7 (Stability of discrete quasi-geodesics. Let l > 0. There exists L = L(l, δ which only depends on δ and l with the following property. Let x 0,..., x m be a sequence of points of X such that (i for every i {1,..., m 1}, x i 1, x i+1 xi l, (ii for every i {1,..., m 2}, x i+1 x i L. Then for all i {0,..., m}, x, x xi l + 5δ. Moreover, for all p X there exists i {0,..., m 1} such that x i+1, x i p x 0, x m p + 2l + 8δ. Proof. Let η > 0. For every i {0,..., m 1} we choose a (1, η-quasi-geodesic γ i joining x i to x i+1. We denote by γ the concatenation γ 0... γ m 1. According to our assumptions this is a L-local (1, 2l + 4η-quasi-geodesic. We can therefore apply Proposition 2.5. If L is sufficiently large then we obtain the followings. (i For every i {0,..., m} we have x, x xi l + 2η + 5δ. (ii For every p X we have d(p, γ x 0, x m p + 2l + 4η + 8δ. However there exists i {0,..., m 1} such that the distance between p and γ is exactly d(p, γ i. The path γ i being a (1, η-quasi-geodesic, we get x i, x i+1 p x 0, x m p + 2l + 5η + 8δ. The inequalities that we obtained hold for every sufficiently small η > 0, which gives the desired conclusion.

10 10 Small cancellation theory and Burnside problem. 2.3 Quasi-convex subsets Definition 2.8. Let α 0. A subset Y of X is α-quasi-convex if for every x X, for every y, y Y, d(x, Y y, y x + α. Remark. Since X is not a geodesic space our definition of quasi-convex slightly differs from the usual one (every geodesic joining two points of Y remains in the α-neighborhood of Y. However if X is geodesic, an α-quasi-convex subset in the usual sense is (α +3δ-quasi-convex in our sense and conversely. According to Proposition 2.4 every (1, l-quasi-geodesic is (l + 3δ-quasi-convex. If L is sufficiently large then every L-local (1, l-quasi-geodesic is (l + 8δ-quasi-convex (see Corollary 2.6. For our purpose we will also need a slightly stronger version of quasi-convexity. Definition 2.9. Let Y be a subset of X connected by rectifiable paths. The length metric on Y induced by the restriction of. X to Y is denoted by. Y. We say that Y is strongly quasi-convex if Y is 2δ-quasi-convex and for every y, y Y, y y X y y Y y y X + 8δ. Remark. The first inequality is just a consequence of the definition of. Y. Lemma 2.10 (Projection on a quasi-convex. Compare [6, Chap. 10, Prop. 2.1]. Let Y be an α-quasi-convex subset of X. Let x, x X. (i If p is an η-projection of x on Y, then for all y Y, x, y p α + η. (ii If p and p are respective η- and η -projections of x and x on Y, then { } p p max x x x p x p + 2ε, ε, where ε = 2α + η + η + δ. Lemma 2.11 (Neighborhood of a quasi-convex. Compare [6, Chap. 10, Prop. 1.2]. Let α 0. Let Y be an α-quasi-convex subset of X. For every A α, the A-neighborhood of Y is 2δ-quasiconvex. Lemma Let Y 1 and Y 2 be respectively α 1 - and α 2 -quasi-convex subsets of X. The subset is 7δ-quasi-convex. Z = Y +α1+3δ 1 Y +α2+3δ 2 Proof. Let x X and t, t Z. Let η (0, 2δ. We denote by γ a (1, η-quasi-geodesic joining t and t. Let u be a point of γ. In particular t, t u η/2. We assume first that t u 4δ + 2η and t u 4δ + 2η. We denote by y and y respective η-projections of t and t on Y i. By hyperbolicity } min { t, y u, y, y u, y, t u t, t u + 2δ 1 η + 2δ. (3 2 Assume that the minimum is achieved by t, y u, it gives y u = t y t u + 2 t, y u α i + 3δ Thus u belongs to the (α i + 3δ-neighborhood of Y i. The same holds if the minimum in (3 is achieved by y, t u. Suppose now that the minimum is achieved by y, y u. The set Y i being α-quasi-convex we get d(u, Y i y, y u + α i α i + 3δ. In all cases u lies in the (α i + 3δ- neighborhood of Y i. It follows that any point of γ is in the (4δ + 2η-neighborhood of Z.

11 2 Hyperbolic geometry 11 Let p be a projection of x on γ. The path γ is (η + 3δ-quasi-convex (Proposition 2.4 hence x p t, t x + η + 3δ. According to the previous remark p lies in the (4δ + 2η-neighborhood of Z thus d(x, Z t, t x + 3η + 7δ. This inequality holds for every sufficiently small η. Thus Z is 7δ-quasi-convex. Lemma Let Y and Z be respectively α- and β-quasi-convex subsets of X. For all A 0 we have diam ( Y +A Z +A diam ( Y +α+3δ Z +β+3δ + 2A + 4δ. Proof. Let x and x be two points of Y +A Z +A. We assume that x x > 2A+4δ. Let η (0, δ such that x x > 2A + 4δ + 6η. There exist t, t X such that x t = x t = A + 2δ + 3η and x, x t, x, x t η. Note that x t, x t A+2δ +3η. We claim that t and t belong to the (α + 3δ-neighborhood of Y. Let us denote by y and y respective η-projection of x and x on Y. By hyperbolicity min { x t x y, y, y t, x t x y } x, x t + 2δ 2δ + η. It follows that y, y t 2δ +η 3δ. The subset Y being α-quasi-convex we get d(t, Y α+3δ. The same holds for t, which proves our claim. Similarly t and t lie in the (β + 3δ-neighborhood of Z. Consequently, x x t t + 2A + 4δ + 6η diam ( Y +α+3δ Z +α+3δ + 2A + 4δ + 6η. This last inequality actually holds for every sufficiently small η and x, x in Y +A Z +A, which leads to the conclusion. Definition Let Y be a subset of X. The hull of Y denoted by hull (Y is the union of all (1, δ-quasi-geodesics joining two points of Y. Lemma Let Y be a subset of X. The hull of Y is 6δ-quasi-convex. Proof. Let x X and y, y hull (Y. By definition there exist γ : [a, b] X and γ : [a, b ] X two (1, δ-quasi-geodesics joining points of Y such that y and y respectively lie on γ and γ. Since X is a length space, there exists a (1, δ-quasi-geodesic γ 0 between γ(a and γ (a. In particular γ 0 hull (Y. By hyperbolicity } min { y, γ(a x, γ(a, γ(a x, γ(a, y x y, y x + 2δ. However γ is 4δ-quasi-convex (Proposition 2.4, thus d(x, hull (Y d(x, γ y, γ(a x + 4δ. We have similar inequalities for γ 0 and γ. Hence d(x, hull (Y y, y x + 6δ. Lemma Let Y and Z be two subsets of X. Let x be a point of X. Assume that for all y Y, for all z Z, y, z x α. Then for all y hull (Y, for all z hull (Z, y, z x α + 3δ. Proof. Let y hull (Y and z hull (Z. By definition there exists y 1, y 2 Y (respectively z 1, z 2 Z such that y (respectively z lies on a (1, δ-quasi-geodesic between y 1 and y 2 (respectively z 1 and z 2. By hyperbolicity } min { y 1, x y, y 2, x y y 1, y 2 y + δ 3 2 δ. In particular there is i {1, 2} such that y i, x y 3δ/2. In the same way there is j {1, 2} such that z j, x z 3δ/2. By triangle inequality we obtain y, z x y i, z j x + y i, x y + z j, x z α + 3δ.

12 12 Small cancellation theory and Burnside problem. 2.4 Ultra-limit of hyperbolic spaces Let us first recall the definition of the ultra-limit of a sequence of metric spaces and some related notations. A non-principal ultra-filter is a finite additive map ω : P(N {0, 1} such that ω(n = 1 and which vanishes on every finite subset of N. A property P n is true ω-almost surely (ω-as if ω ({n N P n is true} = 1. A real sequence (u n is ω-essentially bounded (ω-eb if there exists M such that u n M ω-as. Given l R, we say that the ω-limit of (u n is l and write lim ω u n = l if for all ε > 0, u n l ε ω-as. In particular, any sequence which is ω-eb admits a ω-limit [2]. Let ( X n, x 0 n be a sequence of pointed metric spaces. We define the following set Π ω X n = { ( (x n Π n N X n xn x 0 n is ω-eb }. The space Π ω X n is endowed with a pseudo-metric defined in the following way: (x n (y n = lim ω x n y n. Definition The ω-limit of ( X n, x 0 n, denoted by limω ( Xn, x 0 n or simply limω X n, is the quotient of the space Π ω X n by the equivalence relation which identifies two points at distance zero. The pseudo-distance on Π ω X n induces a distance on lim ω X n. Remark. If the diameter of X n is uniformly bounded, then lim ω ( Xn, x 0 n does not depend on the choice of a base point x 0 n. Notations. Given a sequence (x n Π ω X n we write lim ω x n for its equivalence class in lim ω X n. For all n N, let Y n be a subset of X n. The set lim ω Y n is defined by { } lim Y n = lim y n (yn Π ω X n and y n Y n ω-as. ω ω Proposition 2.18 (Ultra-limit of hyperbolic spaces [[7, Prop ]. ] Let ω be a non-principal ultra-filter. Let ( X n, x 0 n be a sequence of pointed δn -hyperbolic length spaces such that δ = lim ω δ n. Then lim ω X n is a δ-hyperbolic geodesic space. In particular, if δ = 0, lim ω X n is an R-tree. ( Proposition 2.19 ([7, Prop ]. Let δ 0. Let ω be a non-principal ultra-filter. Let Xn, x 0 n be a sequence of pointed length spaces. Assume that limω X n is a δ-hyperbolic space. Then for every η > 0 for every r > 0, every ball of radius of r of X n is (δ + η-hyperbolic ω-as. Proposition Let ω be a non-principal ultra-filter. Let ( X n, x 0 n be a sequence of pointed δ n -hyperbolic length spaces with lim ω δ n = 0. For every n N let Y n and Z n be respectively α n - and β n -quasi-convex subsets of X n. We denote by Y = lim ω Y n and Z = lim ω Z n the corresponding limit subsets of X = lim ω X n. Then diam (Y Z lim ω diam ( Y +αn+3δn n Z n +βn+3δn.

13 2 Hyperbolic geometry 13 Proof. Let x = lim ω x n and x = lim ω x n be two points of Y Z. Let A > 0. Since x and x belong to both Y and Z, x n and x n belong to Y n +A Z n +A ω-as. Applying Proposition 2.13 we obtain x n x n diam ( Y n +αn+3δn Z n +βn+3δn + 2A + 4δn ω-as. After taking the ω-limit, it gives x x lim ω diam ( Y +αn+3δn n Z n +βn+3δn + 2A. This last inequality holds for every x, x Y Z and every A > 0, which leads to the result. 2.5 Isometries of a hyperbolic space In this section we assume that the space X is geodesic and proper. By proper we mean that every closed ball of X is compact. Although it is not necessarily unique, [x, x ] stands for a geodesic between two points x and x of X. We denote by X the boundary at infinity of X (see [6, Chap. 2]. The space X being proper any two distinct points of X are joined by a bi-infinite geodesic. In this situation one can precise the constants that appears in Corollary 2.6 (see [4, Chap. III.H, Th. 1.13]: the Hausdorff distance between two 200δ-local (1, 0-quasi-geodesics of X joining the same extremities (eventually in X is at most 5δ. In particular the Hausdorff distance between two bi-infinite geodesics joining the same points of X is at most 5δ. Lemma Let α 0 and Y be an α-quasi-convex subset of X. For every A α + 2δ, the A-neighborhood of Y is strongly quasi-convex. Remark. An analog statement is true if the space X is not proper of geodesic. However is requires to consider the open A-neighborhood of X. That is why we preferred to state this result with this stronger assumption. Proof. According to Lemma 2.11, it is sufficient to prove that the 2δ-neighborhood of a closed 2δquasi-convex subset Y of X is 8δ-strongly quasi-convex. Let x and x be two points of X which are 2δ-close to Y. We denote by p and p their respective projections on Y. By construction the geodesics [x, p] and [x, p ] lie in the 2δ-neighborhood of Y. By quasi-convexity the same holds for [p, p ]. Thus by concatenating the three geodesics we obtain a path contained in the 2δ-neighborhood of Y joining x to x whose length is at most x x + 8δ. Consequently the 2δ-neighborhood of Y is 8δ-strongly quasi-convex. Let x be a point of X. An isometry g of X is either elliptic i.e., the orbit of x under g is bounded, parabolic i.e., the orbit of x under g has exactly one accumulation point in X. hyperbolic i.e., the orbit of x under g has exactly two accumulation points in X. Note that these definitions do not depend on x. In order to measure the action of g on X, we define two translation lengths. By the translation length [g] X (or simply [g] we mean [g] X = inf gx x. x X The asymptotic translation length [g] X (or simply [g] is [g] X = lim 1 n + n gn x x.

14 14 Small cancellation theory and Burnside problem. These two lengths satisfy the following inequality [g] [g] [g] + 32δ [6, Chap. 10, Prop. 6.4]. An isometry g of X is hyperbolic if and only if [g] > 0 [6, Chap. 10, Prop. 6.3]. Lemma Let x, x and y be three points of X. Let g be an isometry of X. Then gy y max { gx x, gx x } + 2 x, x y + 6δ. Proof. By hyperbolicity } min { x, gx y, gx, gx y, gx, x y x, x y + 2δ. (4 Assume that the minimum is achieved by x, gx y. Using the triangle inequality we obtain gy y gx x + 2 x, gx y gx x + 2 x, x y + 4δ. A similar inequality holds if the minimum is achieved by gx, x y. Suppose now that the minimum in (4 is achieved by gx, gx y. Hence gx, gx y x, x y + 2δ. Applying (2 we obtain { } gy y + gx gx max gx gy + gx y, gx gy + gx y + 2δ. (5 However, by triangle inequality gx gy + gx y gx x + gx gx + 2 gx, gx y gx x + gx gx + 2 x, x y + 4δ. The same inequality holds after swapping x and x. Therefore (5 leads to the desired result. Definition Let g be an isometry of X. The axis of g denoted by A g is the set of points x X such that gx x max{[g], 8δ}. Remarks. Note that we do not require g to be hyperbolic. This definition works also for parabolic or elliptic isometries. This subset is not empty because X is proper. It is also closed. Proposition Let g be an isometry of X. Let x be a point of X. (i gx x 2d(x, A g + [g] 14δ, (ii if gx x [g] + A, then d(x, A g 1 2 A + 7δ, (iii A g is 14δ-quasi-convex. Proof. Let x X. Note that if x belongs to A g, Point (i is true. Therefore we can assume that x / A g. We denote by y a projection of x on A g. Observe that any geodesic [y, gy] is contained in A g. Such a geodesic is 3δ-quasi-convex. Moreover, y and gy are respective projections of x and gx on it. Proposition 2.10 gives { } gy y max gx x 2 x y + 14δ, 7δ. (6 On the other hand we claim that gy y 8δ. Recall that x does not belong to A g. By construction of y any point z on [x, y] distinct from y does not belong to A g, Therefore gz z max{[g], 8δ}. Taking the limit as z approaches y leads to the claim. Hence (6 gives gx x gy y + 2 x y 14δ [g] + 2d(x, A g 14δ, (7 which proves Point (i. Point (ii is a consequence of (i. Let us now prove Point (iii. Let y and y be two points of A g. Let x be a point of X. By Lemma 2.22, { } gx x max gy y, gy y + 2 y, y x + 6δ [g] + 2 y, y x + 14δ. It follows then from Point (ii that d(x, A g y, y x + 14δ.

15 2 Hyperbolic geometry 15 Let g be a hyperbolic isometry of X. We write g and g + for the accumulation points in X of an orbit of g. They are the only points of X fixed by g. Definition Let g be a hyperbolic isometry of X. The cylinder of g denoted by Y g is the set of points lying in the 10δ-neighborhood of a geodesic joining g and g +. Lemma Let g be a hyperbolic isometry of X. The union Γ of all geodesics joining g to g + is 8δ-quasi-convex. The set Y g is strongly quasi-convex. Proof. According to Lemma 2.21 it is sufficient to prove that Γ is 8δ-quasi-convex. Let x X and y, y Γ. There exist γ and γ two geodesics joining g and g + such that y and y respectively lie on γ and γ. We denote by p a projection of y on γ. Since γ and γ join the same extremities the Hausdorff distance between them is at most 5δ. Thus y p 5δ. The path γ being a bi-infinite geodesic, it follows that d(x, Γ d(x, γ y, p x + 3δ y, y x + 8δ. Lemma Let g be a hyperbolic isometry of X. Let Y be a g-invariant α-quasi-convex subset of X. Then Y g lies in the (α + 22δ-neighborhood of Y. In particular Y g is contained in the 36δ-neighborhood of A g. Proof. It is sufficient to prove that every bi-infinite geodesic joining g to g + lies in the (α+12δ- neighborhood of Y. Let γ be such a geodesic and x a point of γ. Let η > 0. We denote by y an η-projection of x on Y. Since g is hyperbolic there exists m N such that g m x g m x > 2 x y + 26δ. The geodesics g m γ and g m γ also join g and g +. Therefore g m x and g m x are 5δ-close to γ. We denote by p and p + respective projections of these points on γ. Note that x lies on the portion of γ between p and p +. Indeed if it was not the case we would have g m x g m x p p δ = x p x p δ 20δ, which contradicts our assumption on m. In particular g m x, g m x x 10δ. By hyperbolicity we get min { g m x x x y, g m y, g m y x} g m x, g m x + 2δ 12δ x By construction of m, g m x x is bounded below by x y + 13δ. Therefore the minimum in the previous inequality is achieved by g m y, g m y x. However Y being g-invariant and α-quasiconvex, g m y and g m y are two points of Y and d(x, Y g m y, g m y x + α. Consequently x lies in the (α + 12δ-neighborhood of Y. Let g be an isometry of X such that [g] > 200δ. (In particular, g is hyperbolic. Let x be a point of A g. We consider a geodesic γ : J X between x and gx parametrized by arc length. We extend γ in a g-invariant path γ : R X in the following way: for all t J, for all m Z, γ (t + m[g] = g m γ(t. This is a [g]-local (1,0-quasi-geodesic contained in A g joining g to g +. By stability of quasi-geodesics γ is actually 8δ-quasi-convex. We call such a path a nerve of g. The Hausdorff distance between two nerves of g is at most 5δ. Lemma Let g G such that [g] > 200δ. Let γ be a 200δ-local (1, 0-quasi-geodesic joining g to g + Then A g is contained in the 5δ-neighborhood of γ. In particular A g is contained in Y g and in the 5δ-neighborhood of any nerve of g. Proof. Let x A g. There exists γ a nerve of g going through x. Both γ and γ are 200δ-local (1, 0-quasi-geodesic joining g to g +. By the stability of quasi-geodesics γ and thus x lies in the 5δ-neighborhood of γ.

16 16 Small cancellation theory and Burnside problem. The next lemma explains the following fact. Let g be a hyperbolic isometry of X. A quasigeodesic contained in the neighborhood of the axis of g almost behaves like a nerve of g. Lemma Let g G such that [g] > 200δ. Let γ : [a, b] X be a [g]-local (1, 0-quasigeodesic contained in the A-neighborhood of A g. Then there exists ε {±1} such that for every s [a, b] if s b [g] then g ε γ(s γ(s + [g] 4A + 80δ. Proof. We denote by γ g an nerve of g. Since [g] > 200δ, the 5δ-neighborhood of γ g contains A g. Thus γ lies in the (A + 5δ-neighborhood of γ g. In particular there exist c, d R such that γ(a γ g (c A + 5δ and γ(b γ g (d A + 5δ. By replacing if necessary g by g 1 we can assume that c d. Let s [a, b] such that s t [g]. Using the stability of quasi-geodesics there exist t [c, d] and t [t, d] such that γ(s γ g (t A + 15δ and γ(s + [g] γ g (t A + 19δ. It follows that γ g(t γ g (t γ(s γ(s + [g] 2A + 34δ. However γ(s γ(s + [g] and γ g (t γ g (t + [g] both equal [g]. Consequently γ g(t γ g (t γ g (t γ g (t + [g] 2A + 34δ. Since t and t + [g] are larger than t we get by Lemma 2.2-(ii. γ g (t gγ g (t = γ g (t γ g (t + [g] 2A + 46δ. It follows then from the triangle inequality that gγ(s γ(s + [g] 4A + 80δ. 2.6 Group acting on a hyperbolic space In this section G denotes a group acting by isometries on X. We still assume that X is geodesic and proper. Moreover, we require the action of G on X to be (i proper i.e., for every x X, there exists r > 0 such that the set of elements g G satisfying gb(x, r B(x, r is finite. (ii co-compact i.e., the quotient X/G endowed with the induced topology is compact. Since the space X is proper, the properness of the action of G implies this more general fact. Let Y be a bounded subset of X. The set of elements g G such that gy intersects Y is finite [4, Chap. I.8, Remark 8.3]. It follows from these assumptions that a subgroup of G is either elementary i.e., virtually cyclic or contains a free group of rank 2 [11, Chap. 8, Th. 37]. Notation. Given a subset Y of X we denote by Stab (Y the stabilizer of Y i.e., the set of elements g G such that gy = Y.

17 2 Hyperbolic geometry 17 Finite subgroups. We start by studying some properties of the finite subgroups of G. To that end we associate to each such subgroup a particular subset of X. Definition Given a finite subgroup F of G we denote by C F the set of points x X such that for every g F, gx x 10δ. It follows from the definition that C F is an F -invariant subset of X. Proposition Let F be a finite subgroup of G. Let x be a point of X. Let g F such that gx x is maximal. We denote by m the midpoint of a geodesic [x, gx]. Then m belongs to C F. In particular C F is non-empty. Proof. Let h be an element of F. In order to simplify the notations we put y = gx, z = hx and t = hgx. The point p = hm is thus the midpoint of [z, t]. The only fact that we are going to use is that x y = z t is the largest distance between any two points of {x, y, z, t}. Using hyperbolicity condition (2 we have x y + z t max { x t + y z, y t + x z } + 2δ. (8 Note that z and t play a symmetric role. Without loss of generality we can assume that the maximum is achieved by x t + y z. It follows that ( ( x y x t + z t y z 2δ. Recall that x y and z t are respectively larger than or equal to x t and y z. Consequently 0 x y x t 2δ. Roughly speaking the triangle [x, y, t] has two sides with approximatively the same length namely [x, y] and [x, t], this length being larger than the one of the last side. It follows that y, t x x m δ. Similarly we have x, z t t p δ. Applying twice Lemma 2.2-(i, we obtain } x, t m max { x m y, t x, x, y m + δ 2δ, } x, t p max { t p x, z t, z, t p + δ 2δ. Lemma 2.2-(ii, leads then to p m x p x m + 2 max { x, t m, x, t p } + 2δ x p x m + 6δ. However m and p are respectively the midpoints of [x, y] and [z, t] which have the same length, thus ( x p x m = x t x y + 2 x, t p. Therefore p m 10δ i.e., hm m 10δ. In other words m belongs to C F. Corollary Let F be a finite subgroup of G. The subset C F is 8δ-quasi-convex. Proof. Let x X and y, y C F. We denote by g an element of F such that gx x is maximal and m the midpoint of [x, gx]. According to Lemma 2.22, 2 x m = gx x max { gy y, gy y } + 2 y, y x + 6δ. However y and y belong to C F, hence x m y, y x + 8δ. By Proposition 2.31, m belongs to C F. Therefore d(x, C F y, y x + 8δ

18 18 Small cancellation theory and Burnside problem. Corollary Let F be a finite subgroup of G. Let Y be a non-empty, F -invariant, α-quasiconvex subset of X. Then the α-neighborhood of Y intersects C F. Proof. Let x be a point of Y. We denote by g an element of F such that gx x is maximal and m the midpoint of [x, gx]. According to Proposition 2.31, m lies in C F. On the other hand, Y is F -invariant, therefore gx Y. Since Y is α-quasi-convex, d(y, Y x, gx m + α α. Consequently m belongs to the α-neighborhood of Y. Infinite elementary subgroups. Let H be an infinite elementary subgroup of G. By definition H contains a finite index subgroup isomorphic to Z. The set of accumulation points in X of an orbit of H, that we denote H, has exactly two points. There exists a subgroup H + of H of index at most 2 which fixes pointwise H. If H + H then H contains an element of order 2. A Schur Theorem (see [25, Th. 5.32] implies that H + contains a unique maximal finite subgroup F. This group is actually a normal subgroup of H +. Moreover, there exists a hyperbolic element h H + so that H + is isomorphic to F Z where Z is identified with the subgroup h acting by conjugation on F. Let g be a hyperbolic element of G. The subgroup E of G that stabilizes {g, g + } is the maximal elementary subgroup of G containing g [6, Chap. 10, Prop. 7.1]. The isometry g is said to be a proper power if there exist h G and an integer n 2 such that g = h n. Any hyperbolic element of G is a power of an isometry which is not a proper power. Lemma Let g be a hyperbolic element of G and H a subgroup of G fixing pointwise {g, g + }. Let F be the maximal finite subgroup of H. The cylinder Y g of g is contained in the 48δ-neighborhood of C F. Proof. According to Lemma 2.26, the union Γ of all geodesics joining g to g + is 8δ-quasi-convex. It is also F -invariant. By Lemma 2.33, there exists a point x in C F Y g. The subgroup F being g invariant, for every n Z, g n x belongs to C F Y g. Note that g n x tends to g + (respectively g and n approaches + (respectively. In particular, for every y Y g there exist n, m Z such that g n x, g m x y 40δ. Since C F is 8δ-quasi-convex, y lies in the 48δ-neighborhood of C F. Lemma Assume that every elementary subgroup of G is cyclic. Let g, h G. If g and hgh 1 generate an elementary subgroup then so do g and h. Proof. We denote by H the subgroup of G generated by g and hgh 1. We distinguish two cases. If g is hyperbolic then g and hgh 1 are two hyperbolic isometries with the same accumulation points in X. In particular h stabilizes the set {g, g + }. It follows that g and h belong to the elementary subgroup E(g. Assume now that g has finite order. The subgroup H is elementary, thus cyclic. In particular it has to be finite. The isometries g and hgh 1 generate two subgroups of H with the same order. However two such subgroups in a cyclic group are equal. Therefore there exists m Z such that hgh 1 = g m. Thus any element of the subgroup generated by g and h can be written g p h q with p, q Z. Since g has finite order, this subgroup is also elementary. Lemma We assume that every elementary subgroup of G is cyclic. Let n N. Let g and h be two hyperbolic elements of G which are not proper powers. Either g and h generate a non-elementary subgroup of G or g n = h n.

19 2 Hyperbolic geometry 19 Proof. Assume that g and h generate an elementary subgroup. This subgroup is infinite and cyclic. Since g and h are not proper powers, they are either equal or inverse. Hence g n = h n. Group invariants. We now introduce several invariants associated to the action of G on X. During the final induction, they will be useful to ensure that the set of relations we are looking at satisfy a small cancellation assumption. Definition Let P be a subset of G. The injectivity radius of P on X, denoted by r inj (P, X is r inj (P, X = inf {[g] g P, hyperbolic} Proposition 2.38 (see [9, Prop. 3.1]. There exists a > 0 such that for every hyperbolic element g G we have [g] an. In particular r inj (G, X > 0. Definition We denote by A the set of pairs (g, h generating a non-elementary subgroup of G such that [g] 1000δ and [h] 1000δ. The parameter A(G, X is given by A(G, X = sup diam ( A +17δ g A +17δ h (g,h A The invariant A(G, X depends implicitly on the hyperbolicity constant δ. Although the notation does not make this dependency explicit, we should keep in mind that it plays an important role. For instance, we have the following lemma: Lemma Let λ be a positive number. We denote by λx the space X endowed with the rescaled metric λ. X and view it as a λδ-hyperbolic space. Then A(G, λx = λa(g, X. Proof. Let g be an element of G. Its translation length satisfies [g] λx = λ[g] X. Since λx is a λδ-hyperbolic space, the axis of g in λx is exactly the image in λx of the axis A g of g in X. We will denote it by λa g. Let g and g be two elements of G that do not generate an elementary subgroup and whose translation lengths in λx are at most than 1000λδ. In particular, we have [g] X, [g ] X 1000δ. By definition of A(G, X, we get ( diam λa +17λδ g λa +17λδ g ( = λ diam A +17δ g A +17δ g λa(g, X. After taking the upper bound for all g and g, we obtain A(G, λx λa(g, X. In the same way, A(G, λx λa(g, X. This establishes the desired equality. Proposition We assume that every elementary subgroup of G is cyclic. Let g and h be two elements of G which generate a non-elementary subgroup. (i If [g] 1000δ, then diam ( A +17δ g A +17δ h [h] + A(G, X + 158δ. (ii Without assumption on g we have, diam ( A +17δ g A +17δ h [g] + [h] + max{[g], [h]} + A(G, X + 676δ. Proof. We prove Point (i by contradiction. Assume that diam ( A +17δ g A +17δ h > [h] + A(G, X + 158δ.